Animated Solution for Mathematics - Trigonometry: If $\cos (\alpha +\beta )=\frac{3}{5},\sin (\alpha -\beta )=\frac{5}{13}$ and $0<\alpha ,\beta <\frac{\pi }{4}$, then $\tan (2\alpha )$ is equal to :

Welcome, future engineer! Today, we are going to look at a problem that might seem like a standard trigonometry exercise, but it is actually a masterclass in pattern recognition.

We are given $\cos (\alpha +\beta )=\frac{3}{5}$ and $\sin (\alpha -\beta )=\frac{5}{13}$, and we are asked to find $\tan (2\alpha )$.

At first glance, you might be tempted to reach for your calculator or start trying to isolate $\alpha$ and $\beta$. Stop! Take a breath. In the world of JEE, the most elegant path is rarely the brute-force one.

The secret here lies in the relationship between the angles. Look at the target: $2\alpha$.

Now, look at the given angles: $(\alpha +\beta )$ and $(\alpha -\beta )$. If we add these two compound angles together, the $\beta$ terms cancel out perfectly:

This is the 'Aha!' moment. We don't need to know $\alpha$ or $\beta$ individually; we just need to treat them as building blocks. We can now write $\tan (2\alpha )$ as $\tan ((\alpha +\beta )+(\alpha -\beta ))$.

Now that we have our strategy, we need our tools. To use the tangent addition formula, we need the tangent of our two building blocks.

Let's start with $\cos (\alpha +\beta )=\frac{3}{5}$. Imagine a right-angled triangle where the base is $3$ and the hypotenuse is $5$. By the Pythagorean theorem, the perpendicular must be $\sqrt{5^2-3^2}=4$.

Thus, we find:

Next, let's look at $\sin (\alpha -\beta )=\frac{5}{13}$. Construct another triangle where the perpendicular is $5$ and the hypotenuse is $13$. The base is $\sqrt{13^2-5^2}=12$.

So, we have:

Now, we apply the tangent addition formula: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Setting $A=\alpha +\beta$ and $B=\alpha -\beta$, we substitute our values:

Let's handle the numerator:

Now the denominator:

Finally, we divide the numerator by the denominator:

Simplifying this, $12$ goes into $36$ three times, leaving us with:

The beauty of this problem is not just in the final number, but in the realization that by simply rearranging the components, we turned a complex-looking problem into a straightforward calculation. Keep looking for these connections, and you will find that math becomes much more than just equations—it becomes a language of logic. You have got this!